Question

A rescue plane spots a person floating in a lifeboat​ 55 m directly below and releases an emergency kit with a parachute. The package descends with a constant vertical acceleration of 6.91 m/s2 (think that’s ^2). If the horizontal speed of the plane was 70.6 m/s, how far is the package from the lifeboat when it hits the waves?

Answer 1

Answer:

281.7m

Explanation:

Given: A lifeboat is 55m directly below a rescue plane, constant vertical acceleration of 6.91 $\text{m}\setminus\text{s}^2$, horizontal speed of plane is $70.6 \text{m}\setminus\text{s}$

To Find: how far is the package from the lifeboat when it hits the waves

Solution: Verticle distance to cover=$55\text{m}$

The emergency kit descends with acceleration of $6.91 \text{m}\setminus\text{s}^2$

Using second equation of motion

$\text{S}=\text{ut}+\frac{1}{2}\text{a}\text{t}^2$

initial speed of emergency kit$(\text{u})=0$

$\text{a}=6.91 \text{m}\setminus\text{s}^2$

putting in equation

$55=0\text{t}+\frac{1}{2}6.91\text{t}^2$

$\text{t}^2=\frac{55\times2}{6.91}$

$\text{t}^2=15.92$

$\text{t}=3.99\text{s}$

time taken by kit to hit the wave$3.99\text{s}$

distace of kit from lifeboat=horizontal distance travelled by emergency kit

horizontal velocity of kit will be same as that of plane as kit is thrown from plane

horizontal distance travelled by emergency kit

using equation of motion

$\text{S}=\text{ut}$

here $\text{u}=70.6\text{m}\setminus\text{s}$

$\text{S}=70.6\times3.99$

$\text{S}=281.69\text{m}$≅$281.7\text{m}$

The distance of emergency kit from lifeboat is $281.7\text{m}$

Answer 2

55 = (1/2) 6.91 t^2

t = about 4

4 * 71 = 284